Question

show the area of a square is equal to half the product it's diagonal​

Answer 1

Let ABCD is a square, each side is x unit. Diagonals AC =BD =y unit.If diagonals intersect at point O. Angle AOB=90° and OA=OB= y/2.

In right angled triangle AOB

OA^2+OB^2=AB^2

(y^2)/4+(y^2)/4=x^2 or x^2=(y^2)/2…………..(1)

Area of square=(side)^2=(x)^2 , [put x^2=(y^2)/2 from eq.(1).]

Area of square =(y^2)/2=(1/2)×y×y=(1/2)AC×BD.

= Half of the product of diagonals.

Hence Proved(◕ᴗ◕✿)

hope it helps you ( ꈍᴗꈍ)

Answer 2

Answer:

Area of a square = Half of the product of diagonals.

Step-by-step explanation:

Let ABCD be a square with each side of s units.

Diagonals AC = BD = d unit. If diagonals intersect at point O, angle AOB=90° and OA=OB= d/2.

In right angled triangle AOB,

⇒ $OA^{2}$+$OB^{2}$=$AB^{2}$

⇒$\frac{d^{2}}{4}$ + $\frac{ d^{2} }{4}$ = $s^{2}$

Area of square= $side^{2}$ = $s^{2}$  

                        = $\frac{d^{2}}{2}$= $\frac{1}{2}$ × d × d = $\frac{1}{2}$ × AC × BD

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